Answer:
The instantaneous velocity of the rocket the moment before it hits the ground is 50 m/s.
Explanation:
Given;
initial velocity of the rocket, u = 50 m/s
Determine the maximum height reached by the rocket.
at maximum height reached by the rocket, the final velocity, v = 0
v² = u² -2gh
0 = 50² - 2(9.8)h
19.6h = 2500
h = 2500 / 19.6
h = 127.55 m
At maximum height, the time to reach ground is given by;
h = ¹/₂gt²
Before the rocket hits the ground the final velocity will be maximum;
v = u + gt
v = 0 + 9.8 x 5.1
v = 50 m/s
Therefore, the instantaneous velocity of the rocket the moment before it hits the ground is 50 m/s.
<u>Answer:</u>
Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s
Direction of resultant velocity of kayaker = 49.32⁰ South of west.
<u>Explanation:</u>
Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.
First kayaker paddles at 4.0 m/s in a direction 30° south of west, kayaker paddles at 4.0 m/s in a direction 210° anticlockwise from positive horizontal axis.
So velocity of kayaker = 4 cos 210 i + 4 sin 210 j = -3.46 i - 2 j
He then turns and paddles at 3.7 m/s in a direction 20° west of south, kayaker paddles at 3.7 m/s in a direction 250° anticlockwise from positive horizontal axis.
So that velocity = -1.27 i - 3.48 j
So resultant velocity of kayaker = -3.46 i - 2 j +(-1.27 i - 3.48 j) = -4.71 i - 5.48 j
Magnitude of resultant velocity of kayaker = 
Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s
Direction of resultant positive horizontal axis, θ = tan⁻¹(-5.48/-4.71) = 229.32⁰ = 49.32⁰ South of west.
Watch this ! You've never seen anything like it !
No wait. I'm pretty sure you have.
Weight = (mass) · (gravity)
Weight = (10 kg) · (9.8 m/s²)
Weight = (10 · 9.8) kg·m/s²
Weight = 98 Newtons
The can will burst if you do
This is a problem because due to human actions, we have produced so much of it, it has become dangerous.
Hope this helped!
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